Various types of LiDAR, radar, and sonar have been developed to measure range. In addition to measuring distance, LiDAR has been utilized to measure the concentration of gases or other substances in the atmosphere. For example, LiDAR systems have been mounted in artificial satellites in orbit above the earth. Known differential absorption LiDAR transmits laser light with two wavelengths, and receives reflected light from the surface of the planet and/or gases or other substances in the atmosphere. An example of a known LiDAR technology is described in “Coherent Laser Radar Transceiver for the NASA CO2 Laser Absorption Spectrometer Instrument” (edited by M. W. Phillips et al., Proceedings of 13th Coherent Laser Radar Conference, 2005, pp. 118-121).
In differential absorption LiDAR systems, one of the two wavelengths is selected to have a relatively large absorption coefficient with respect to a specific gas to be measured, and the other wavelength is selected to have a relatively small absorption coefficient. Due to the differences in absorption coefficients, the amount of light that returns to the LiDAR detector depends upon the concentration of the gas or other substance to be measured. The concentration of the gas to be measured is determined from the difference between the amounts of light at the intensity of light detected at the two wavelengths.
Pseudorandom noise (PN) codes have been utilized in LiDAR and radar to measure distances. In general, a PN signal includes a sequence of pulses. The exact same sequence can be generated at both a transmitter and a receiver, such that the sequence generated by the receiver has a very high correlation with the transmitted sequence. However, known systems suffer from various drawbacks.
It is known that an AC coupled signal can be represented by:S=η+α pnΔ  (1.0)where η is the noise, α is the attenuation factor and:pniΔ=2PNi′Δ−1   (2.0)where PN is the PN code represented by 1's and 0's and A represents the phase shift. The standard deviation of the noise can be represented by:
                    σ        =                                            ∑                              i                =                1                            MN                        ⁢                                                            (                                                            η                      i                                        -                                          η                      ave                                                        )                                2                            MN                                                          (        3.0        )            where we have summed over the total code length of the code, where M is the samples per code bit and N is the code bit length. After performing the correlation calculation, it is already known that the signal strength increases by a factor of MN where they are correlated. The correlation function of the total signal is:
                                                                        R                ⁡                                  (                  j                  )                                            =                            ⁢                                                ∑                                      i                    =                    1                                    MN                                ⁢                                                      (                                                                  η                        i                                            +                                              α                        ⁢                                                                                                  ⁢                                                  pn                          i                          Δ                                                                                      )                                    ⁢                                      pn                                          i                      +                      j                                        0                                                                                                                          =                            ⁢                                                                    ∑                                          i                      =                      1                                        MN                                    ⁢                                                            η                      i                                        ⁢                                          pn                                              i                        +                        j                                            0                                                                      +                                  α                  ⁢                                                            ∑                                              i                        =                        1                                            MN                                        ⁢                                                                  pn                        i                        Δ                                            ⁢                                              pn                                                  i                          +                          j                                                0                                                                                                                                                                    =                            ⁢                                                n                  j                  ′                                +                                  S                  j                  ′                                                                                        (        4.0        )            
The standard deviation of the transformed noise η′ is assumed to have 0 mean (AC coupled). The variance can be determined by the equation:
                              (                      σ            ′                    )                =                              〈                                          (                                  η                  ′                                )                            2                        〉                    =                                    1              MN                        ⁢                                          ∑                                  j                  =                  1                                MN                            ⁢                                                (                                      η                    j                    ′                                    )                                2                                                                        (        5.0        )            
To find the sum:
                                          (                          η              j              ′                        )                    2                =                                            ∑                              m                =                1                            MN                        ⁢                                          η                m                            ⁢                              pn                                  m                  +                  j                                0                            ⁢                                                ∑                                      n                    =                    1                                    MN                                ⁢                                                      η                    n                                    ⁢                                      pn                                          n                      +                      j                                        0                                                                                =                                    ∑                              n                ,                m                            MN                        ⁢                                          η                m                            ⁢                              η                n                            ⁢                              pn                                  m                  +                  j                                0                            ⁢                              pn                                  n                  +                  j                                0                                                                        (        6.0        )            so that:
                                          (                          σ              ′                        )                    2                =                              1            MN                    ⁢                                    ∑                              j                =                1                            MN                        ⁢                                          ∑                                  n                  ,                  m                                MN                            ⁢                                                η                  m                                ⁢                                  η                  n                                ⁢                                  pn                                      m                    +                    j                                    0                                ⁢                                  pn                                      n                    +                    j                                    0                                                                                        (        7.0        )            
If the noise has 0 mean, the only part of the sum that will contribute is when n=m so:
                                                                                          (                                      σ                    ′                                    )                                2                            =                            ⁢                                                1                  MN                                ⁢                                                      ∑                                          j                      =                      1                                        MN                                    ⁢                                                            ∑                      m                      MN                                        ⁢                                                                                            η                          m                          2                                                ⁡                                                  (                                                      pn                                                          m                              +                              j                                                        0                                                    )                                                                    2                                                                                                                                              =                            ⁢                                                1                  MN                                ⁢                                                      ∑                                          j                      =                      1                                        MN                                    ⁢                                                            ∑                      m                      MN                                        ⁢                                          η                      m                      2                                                                                                                                              =                            ⁢                                                1                  MN                                ⁢                                                      ∑                                          j                      =                      1                                        MN                                    ⁢                                      MN                    ⁢                                                                                  ⁢                                          σ                      2                                                                                                                                              =                            ⁢                              MN                ⁢                                                                  ⁢                                  σ                  2                                                                                        (        8.0        )            
In the above since pn is either a 1 or −1, its square is always 1. The result is that:σ′=√{square root over (MN)}σ  (9.0)
The new correlated snr in terms of the original snr is:
                              snr          ′                =                                            NM              ⁢                                                          ⁢              α                                                      MN                            ⁢              σ                                =                                    MN                        ⁢            snr                                              (        10.0        )            
Expressed differently in terms of previously defined variables:
                                          snr            ′                    snr                =                              MN                    =                                                    samplerate                codebitrate                            ⁢              N                                                          (        11.0        )            
This may be used in combination range and resolution equations in order to optimize the system.
In the event the signal processed with P codes back to back it is easy to verify that:
                                          snr            ′                    snr                =                              MNP                    =                                                    samplerate                codebitrate                            ⁢              NP                                                          (        12.0        )            
This is useful because it says an entire block of data may be processed at once to get an averaging effect, and the signal to noise can be increased by simply taking a lot of data.
For noise with a 1/f power spectrum, it has been verified numerically that:
                                                        snr              ′                        snr                    ≥                      MNP                          =                                            samplerate              codebitrate                        ⁢            NP                                              (        13.0        )            